LeetCode 303 Range Sum Query - Immutable

LeetCode

题目描述

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

分析

题目中描述，函数可能被调用很多次，并且函数是不可变的。那么可以构造一个数组sum，sum[i] 存放数组 nums从 0 到 i 的和，这样能在 O(1) 的复杂度内求出指定范围的和。

代码

public class NumArray {
    public long[] sum;
    public NumArray(int[] nums) {
        sum = new long[nums.length + 1];
        for (int i = 0; i < nums.length; i++) {
            sum[i + 1] += nums[i] + sum[i];
        }
    }
    public int sumRange(int i, int j) {
        if (i > j || i < 0 || j >= sum.length - 1) {
            return Integer.MIN_VALUE;
        }
        return (int) (sum[j + 1] - sum[i]);
    }
}
// Your NumArray object will be instantiated and called as such:
// NumArray numArray = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);